A recent article in the Triangle Business Journal discusses a firm called Strata Solar. The firm has installed a 900 kilowatt solar power plant near Boone.
As is common with stories on renewable energy, there’s an attempt to try and equate the size of the renewable energy plant (be it wind, solar, etc.) with the amount of homes it will provide power for.
In this article, it says the 900 kilowatts of solar is “sufficient to power about 800 homes.”
These comparisons are misleading because they suggest, for example as in this case, that those 800 homes could rely exclusively on solar. If solar was relied upon exclusively, then those homes would be in the dark most of the time.
There’s an even bigger issue though in this article, which I admit even the renewable energy proponents tend not to mislead about (usually).
When determining how many homes a renewable energy plant (or any energy facility for that matter) can power, one can’t simply look at the installed capacity of a plant and divide that number by how much electricity a typical home consumes.
This article does precisely that. It isn’t clear whether it simply takes 900 kilowatts and divides it by 1.1 kilowatts, which it says is the power typically required to power a single home for a year.
Energy use needs to be discussed in kilowatt-hours, not kilowatts. 900 kilowatts divided by 1.1 kilowatts is 818 homes, but as mentioned, the calculation needs to be done based on kilowatts-hours (which is a measure of actual energy generation, not energy capacity or power).
First, I’ll take the capacity of the solar facility (900 kilowatts or .9 megawatts) and multiply that by the number of hours in a year (8,760). This gets me 7,884 megawatt-hours (I multiplied .9 * 8,760). To convert this number to kilowatt hours, I simply add three zeroes (7,884,000).
According to the Energy Information Administration, the average electricity usage by a household is about 11,000 kilowatt-hours per year.
To figure out how many homes the solar plant could provide electricity for (without taking into account a critical factor as I’ll discuss), I simply divide 7,884,000 kilowatt-hours by 11,000 kilowatt-hours. This gives me about 717 homes.
This number is still inaccurate because it forgets about capacity factor.
Capacity factor, in very simple terms, is the ratio of net electricity generated compared to the maximum electricity that could have been generated if a plant is always at full power. For example, in a year, if a plant at maximum output could generate 1,000 kilowatt-hours and in reality it generates 250 kilowatt-hours, the capacity factor is 25%.
While I consider the number to be unrealistic and inconsistent with actual observations (the number is closer to 15-16 percent), the EIA assumes solar’s capacity factor to be at 25 percent. Progress Energy has estimated the capacity factor in NC to be about 16 percent.
To provide a more accurate estimate of the number of homes, we must take 7,884,000 and multiply that by 25 percent. This gives me 1,971,000 kilowatt hours, which is the amount of electricity that would be generated.
When I take 1,971,000 and divide that by 11,000 kilowatt hours, I get 179 homes.
So, even using the misleading home comparison, this solar power plant using the very generous 25 percent capacity factor would generate electricity for 179 homes, not 800 homes. That’s a big difference.
Plus, these 179 homes, as I indicated, couldn’t rely on solar alone. The homes would usually be in the dark because the sun doesn’t always shine. Even when solar could generate electricity, there would be a need for back-up generation because of solar’s unreliability.